# kappa-Dowker space and the first conjecture of Morita

Recall the product space of the Michael line and the space of the irrational numbers. Even though the first factor is a normal space (in fact a paracompact space) and the second factor is a metric space, their product space is not normal. This is one of the classic examples demonstrating that normality is not well behaved with respect to product space. This post presents an even more striking result, i.e., for any non-discrete normal space $Y$, there exists another normal space $X$ such that $X \times Y$ is not normal. The example of the non-normal product of the Michael line and the irrationals is not some isolated example. Rather it is part of a wide spread phenomenon. This result guarantees that no matter how nice a space $Y$ is, a counter part $X$ can always be found that the product of the two spaces is not normal. This result is known as Morita’s first conjecture and was proved by Atsuji and Rudin. The solution is based on a generalization of Dowker’s theorem and a construction done by Rudin. This post demonstrates how the solution is put together.

All spaces under consideration are Hausdorff.

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Morita’s First Conjecture

In 1976, K. Morita posed the following conjecture.

Morita’s First Conjecture
If $Y$ is a normal space such that $X \times Y$ is a normal space for every normal space $X$, then $Y$ is a discrete space.

The proof given in this post is for proving the contrapositive of the above statement.

Morita’s First Conjecture
If $Y$ is a non-discrete normal space, then there exists some normal space $X$ such that $X \times Y$ is not a normal space.

Though the two forms are logically equivalent, the contrapositive form seems to have a bigger punch. The contrapositive form gives an association. Each non-discrete normal space is paired with a normal space to form a non-normal product. Examples of such pairings are readily available. Michael line is paired with the space of the irrational numbers (as discussed above). The Sogenfrey line is paired with itself. The first uncountable ordinal $\omega_1$ is paired with $\omega_1+1$ (see here) or paired with the cube $I^I$ where $I=[0,1]$ with the usual topology (see here). There are plenty of other individual examples that can be cited. In this post, we focus on a constructive proof of finding such a pairing.

Since the conjecture had been affirmed positively, it should no longer be called a conjecture. Calling it Morita’s first theorem is not appropriate since there are other results that are identified with Morita. In this discussion, we continue to call it a conjecture. Just know that it had been proven.

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Dowker’s Theorem

Next, we examine Dowker’s theorem, which characterizes normal countably paracompact spaces. The following is the statement.

Theorem 1 (Dowker’s Theorem)
Let $X$ be a normal space. The following conditions are equivalent.

1. The space $X$ is countably paracompact.
2. Every countable open cover of $X$ has a point-finite open refinement.
3. If $\left\{U_n: n=1,2,3,\cdots \right\}$ is an open cover of $X$, there exists an open refinement $\left\{V_n: n=1,2,3,\cdots \right\}$ such that $\overline{V_n} \subset U_n$ for each $n$.
4. The product space $X \times Y$ is normal for any compact metric space $Y$.
5. The product space $X \times [0,1]$ is normal where $[0,1]$ is the closed unit interval with the usual Euclidean topology.
6. For each sequence $\left\{A_n \subset X: n=1,2,3,\cdots \right\}$ of closed subsets of $X$ such that $A_1 \supset A_2 \supset A_3 \supset \cdots$ and $\cap_n A_n=\varnothing$, there exist open sets $B_1,B_2,B_3,\cdots$ such that $A_n \subset B_n$ for each $n$ such that $\cap_n B_n=\varnothing$.

The theorem is discussed here and proved here. Any normal space that violates any one of the conditions in the theorem is said to be a Dowker space. One such space was constructed by Rudin in 1971 [2]. Any Dowker space would be one factor in a non-normal product space with the other factor being a compact metric space. Actually much more can be said.

The Dowker space constructed by Rudin is the solution of Morita’s conjecture for a large number of spaces. At minimum, the product of any infinite compact metric space and the Dowker space is not normal as indicated by Dowker’s theorem. Any nontrivial convergent sequence plus the limit point is a compact metric space since it is homeomorphic to $S=\left\{0 \right\} \cup \left\{\frac{1}{n}: n=1,2,3,\cdots \right\}$ (as a subspace of the real line). Thus Rudin’s Dowker space has non-normal product with $S$. Furthermore, the product of Rudin’s Dowker space and any space containing a copy of $S$ is not normal.

Spaces that contain a copy of $S$ extend far beyond the compact metric spaces. Spaces that have lots of convergent sequences include first countable spaces, Frechet spaces and many sequential spaces (see here for an introduction for these spaces). Thus any Dowker space is an answer to Morita’s first conjecture for the non-discrete members of these classes of spaces. Actually, the range for the solution is wider than these spaces. It turns out that any space that has a countable non-discrete subspace would have a non-normal product with a Dowker space. These would include all the classes mentioned above (first countable, Frechet, sequential) as well as countably tight spaces and more.

Therefore, any Dowker space, a normal space that is not countably paracompact, is severely lacking in ability in forming normal product with another space. In order to obtain a complete solution to Morita’s first conjecture, we would need a generalized Dowker’s theorem.

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Shrinking Properties

The key is to come up with a generalized Dowker’s theorem, a theorem like Theorem 1 above, except that it is for arbitrary infinite cardinality. Then a $\kappa$-Dowker space is a space that violates one condition in the theorem. That space would be a candidate for the solution of Morita’s first conjecture. Note that Theorem 1 is for the infinite countable cardinal $\omega$ only. Before stating the theorem, let’s gather all the notions that will go into the theorem.

Let $X$ be a space. Let $\mathcal{U}$ be an open cover of the space $X$. The open cover $\mathcal{U}$ is said to be shrinkable if there is an open cover $\mathcal{V}=\left\{V(U): U \in \mathcal{U} \right\}$ such that $\overline{V(U)} \subset U$ for each $U \in \mathcal{U}$. When this is the case, the open cover $\mathcal{V}$ is said to be a shrinking of $\mathcal{U}$. If an open cover is shrinkable, we also say that the open cover can be shrunk (or has a shrinking).

Let $\kappa$ be a cardinal. The space $X$ is said to be a $\kappa$-shrinking space if every open cover of cardinality $\le \kappa$ of the space $X$ is shinkable. The space $X$ is a shrinking space if it is a $\kappa$-shrinking space for every cardinal $\kappa$.

When a family of sets are indexed by ordinals, the notion of an increasing or decreasing family of sets is possible. For example, the family $\left\{A_\alpha \subset X: \alpha<\kappa \right\}$ of subsets of the space $X$ is said to be increasing if $A_\beta \subset A_\gamma$ whenever $\beta<\gamma$. In other words, for an increasing family, the sets are getting larger whenever the index becomes larger. A decreasing family of sets is defined in the reverse way. These two notions are important for some shrinking properties discussed here – e.g. using an open cover that is increasing or using a family of closed sets that is decreasing.

In the previous discussion on shrinking spaces, two other shrinking properties are discussed – property $\mathcal{D}(\kappa)$ and property $\mathcal{B}(\kappa)$. A space $X$ is said to have property $\mathcal{D}(\kappa)$ if every increasing open cover of cardinality $\le \kappa$ for the space $X$ is shrinkable. A space $X$ is said to have property $\mathcal{B}(\kappa)$ if every increasing open cover of cardinality $\le \kappa$ for the space $X$ has a shrinking that is increasing. See this previous post for a discussion on property $\mathcal{D}(\kappa)$ and property $\mathcal{B}(\kappa)$.

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An Attempt for a Generalized Dowker’s Theorem

Let $\kappa$ be an infinite cardinal. The space $X$ is said to be a $\kappa$-paracompact space if every open cover $\mathcal{U}$ of $X$ with $\lvert \mathcal{U} \lvert \le \kappa$ has a locally finite open refinement. Thus a space is paracompact if it is $\kappa$-paracompact for every infinite cardinal $\kappa$. Of course, an $\omega$-paracompact space is a countably paracompact space.

For any infinite $\kappa$, let $D_\kappa$ be a discrete space of size $\kappa$. Let $p$ be a point not in $D_\kappa$. Define the space $Y_\kappa=D_\kappa \cup \left\{p \right\}$ as follows. The subspace $D_\kappa$ is discrete as before. The open neighborhoods at $p$ are of the form $\left\{ p \right\} \cup B$ where $B \subset D_\kappa$ and $\lvert D_\kappa-B \lvert<\kappa$. In other words, any open set containing $p$ contains all but less than $\kappa$ many discrete points.

Another concept that is needed is the cardinal function called minimal tightness. Let $Y$ be any space. Define the minimal tightness $mt(Y)$ as the least infinite cardinal $\kappa$ such that there is a non-discrete subspace of $Y$ of cardinality $\kappa$. If $Y$ is a discrete space, then let $mt(Y)=0$. For any non-discrete space $Y$, $mt(Y)=\kappa$ for some infinite $\kappa$. Note that for the space $Y_\kappa$ defined above would have $mt(Y_\kappa)=\kappa$. For any space $Y$, $mt(Y)=\omega$ if and only if $Y$ has a countable non-discrete subspace.

The following theorem can be called a $\kappa$-Dowker’s Theorem.

Theorem 2
Let $X$ be a normal space. Let $\kappa$ be an infinite cardinal. Consider the following conditions.

1. The space $X$ is a $\kappa$-paracompact space.
2. The space $X$ is a $\kappa$-shrinking space.
• For each open cover $\left\{U_\alpha: \alpha<\kappa \right\}$ of $X$, there exists an open cover $\left\{V_\alpha: \alpha<\kappa \right\}$ such that $\overline{V_\alpha} \subset U_\alpha$ for each $\alpha<\kappa$.
3. The space $X$ has property $\mathcal{D}(\kappa)$.
• For each increasing open cover $\left\{U_\alpha: \alpha<\kappa \right\}$ of $X$, there exists an open cover $\left\{V_\alpha: \alpha<\kappa \right\}$ such that $\overline{V_\alpha} \subset U_\alpha$ for each $\alpha<\kappa$.
4. For each decreasing family $\left\{F_\alpha: \alpha<\kappa \right\}$ of closed subsets of $X$ such that $\bigcap_{\alpha<\kappa} F_\alpha=\varnothing$, there exists a family $\left\{G_\alpha: \alpha<\kappa \right\}$ of open subsets of $X$ such that $\bigcap_{\alpha<\kappa} G_\alpha=\varnothing$ and $F_\alpha \subset G_\alpha$ for each $\alpha<\kappa$.
5. The space $X$ has property $\mathcal{B}(\kappa)$.
• For each increasing open cover $\left\{U_\alpha: \alpha<\kappa \right\}$ of $X$, there exists an increasing open cover $\left\{V_\alpha: \alpha<\kappa \right\}$ such that $\overline{V_\alpha} \subset U_\alpha$ for each $\alpha<\kappa$.
6. The product space $X \times Y_\kappa$ is a normal space.
7. The product space $X \times Y$ is a normal space for some space $Y$ with $mt(Y)=\kappa$.

The following diagram shows how these conditions are related.

Diagram 1
$\displaystyle \begin{array}{ccccc} 1 &\text{ } & \Longrightarrow & \text{ } & 5 \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ \Downarrow & \text{ } & \text{ } & \text{ } & \Updownarrow \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ 2 &\text{ } & \text{ } & \text{ } & 6 \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ \Downarrow & \text{ } & \text{ } & \text{ } & \Downarrow \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ 3 & \text{ } & \Longleftarrow & \text{ } & 7 \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ \Updownarrow & \text{ } & \text{ } & \text{ } & \text{ } \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ 4 & \text{ } & \text{ } & \text{ } & \text{ } \end{array}$

In addition to Diagram 1, we have the relations $5 \Longrightarrow 3$ and $2 \not \Longrightarrow 5$.

Remarks
At first glance, Diagram 1 might give the impression that the conditions in the theorem form a loop. It turns out the strongest property is $\kappa$-paracompactness (condition 1). Since condition 2 does not imply condition 5, condition 2 does not imply condition 1. Thus the conditions do not form a loop.

The implications $1 \Longrightarrow 2 \Longrightarrow 3 \Longleftarrow 5$ and $6 \Longrightarrow 7$ are immediate. The following implications are established in this previous post.

$3 \Longleftrightarrow 4$ (Theorem 4)

$5 \Longleftrightarrow 6$ (Theorem 7)

$2 \not \Longrightarrow 5$ (Example 1)

The remaining implications to be shown are $1 \Longrightarrow 5$ and $7 \Longrightarrow 3$.

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Proof of Theorem 2

$1 \Longrightarrow 5$
Let $\mathcal{U}=\left\{U_\alpha: \alpha<\kappa \right\}$ be an increasing open cover of $X$. By $\kappa$-paracompactness, let $\mathcal{G}$ be a locally finite open refinement of $\mathcal{U}$. For each $\alpha<\kappa$, define $W_\alpha$ as follows:

$W_\alpha=\cup \left\{G \in \mathcal{G}: G \subset U_\alpha \right\}$

Then $\mathcal{W}=\left\{W_\alpha: \alpha<\kappa \right\}$ is still a locally finite refinement of $\mathcal{U}$. Since the space $X$ is normal, any locally finite open cover is shrinkable. Let $\mathcal{E}=\left\{E_\alpha: \alpha<\kappa \right\}$ be a shrinking of $\mathcal{W}$. The open cover $\mathcal{E}$ is also locally finite. For each $\alpha$, let $V_\alpha=\bigcup_{\beta<\alpha} E_\beta$. Then $\mathcal{V}=\left\{V_\alpha: \alpha<\kappa \right\}$ is an increasing open cover of $X$. Note that

$\overline{V_\alpha}=\overline{\bigcup_{\beta<\alpha} E_\beta}=\bigcup_{\beta<\alpha} \overline{E_\beta}$

since $\mathcal{E}$ is locally finite and thus closure preserving. Since $\mathcal{U}$ is increasing, $\overline{E_\beta} \subset W_\beta \subset U_\beta \subset U_\alpha$ for all $\beta<\alpha$. This means that $\overline{V_\alpha} \subset U_\alpha$ for all $\alpha$.

$7 \Longrightarrow 3$
Since condition 3 is equivalent to condition 4, we show $7 \Longrightarrow 4$. Suppose that $X \times Y$ is normal where $Y$ is a space such that $mt(Y)=\kappa$. Let $D=\left\{d_\alpha: \alpha<\kappa \right\}$ be a non-discrete subset of $Y$. Let $p$ be a point such that $p \ne d_\alpha$ for all $\alpha$ and such that $p$ is a limit point of $D$ (this means that every open set containing $p$ contains some $d_\alpha$). Let $\mathcal{F}=\left\{F_\alpha: \alpha<\kappa \right\}$ be a decreasing family of closed subsets of $X$ such that $\bigcap_{\alpha<\kappa} F_\alpha=\varnothing$. Define $H$ and $K$ as follows:

$H=\cup \left\{F_\alpha \times \left\{d_\alpha \right\}: \alpha<\kappa \right\}$

$K=X \times \left\{p \right\}$

The sets $H$ and $K$ are clearly disjoint. The set $K$ is clearly a closed subset of $X \times Y$. To show that $H$ is closed, let $(x,y) \in (X \times Y)-H$. Two cases to consider: $x \in F_0$ or $x \notin F_0$ where $F_0$ is the first closed set in the family $\mathcal{F}$.

The first case $x \in F_0$. Let $\beta<\kappa$ be least such that $x \notin F_\beta$. Then $y \ne d_\gamma$ for all $\gamma<\beta$ since $(x,y) \in (X \times Y)-H$. In the space $Y$, any subset of cardinality $<\kappa$ is a closed set. Let $E=Y-\left\{d_\gamma: \gamma<\beta \right\}$, which is open containing $y$. Let $O \subset X$ be open such that $x \in O$ and $O \cap F_\beta=\varnothing$. Then $(x,y) \in O \times E$ and $O \times E$ misses points of $H$.

Now consider the second case $x \notin F_0$. Let $O \subset X$ be open such that $x \in O$ and $O$ misses $F_0$. Then $O \times Y$ is an open set containing $(x,y)$ such that $O \times Y$ misses $H$. Thus $H$ is a closed subset of $X \times Y$.

Since $X \times Y$ is normal, choose open $V \subset X \times Y$ such that $H \subset V$ and $\overline{V} \cap K=\varnothing$. For each $\alpha<\kappa$, define $G_\alpha$ as follows:

$G_\alpha=\left\{x \in X: (x,d_\alpha) \in V \right\}$

Note that each $G_\alpha$ is open in $X$ and that $F_\alpha \subset G_\alpha$ for each $\alpha<\kappa$. We claim that $\bigcap_{\alpha<\kappa} G_\alpha=\varnothing$. Let $x \in X$. The point $(x,p)$ is in $K$. Thus $(x,p) \notin \overline{V}$. Choose an open set $L \times M$ such that $(x,p) \in L \times M$ and $(L \times M) \cap \overline{V}=\varnothing$. Since $p \in M$, there is some $\gamma<\kappa$ such that $d_\gamma \in M$. Since $(x,d_\gamma) \notin \overline{V}$, $(x,d_\gamma) \notin V$. Thus $x \notin G_\gamma$. This establishes the claim that $\bigcap_{\alpha<\kappa} G_\alpha=\varnothing$.

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$\kappa$-Dowker Space

Analogous to the Dowker space, a $\kappa$-Dowker space is a normal space that violates one condition in Theorem 2. Since the seven conditions listed in Theorem 7 are not all equivalent, which condition to use? Condition 1 is the strongest condition since it implies all the other condition. At the lower left corner of Diagram 1 is condition 3, which follows from every other condition. Thus condition 3 (or 4) is the weakest property. An appropriate definition of a $\kappa$-Dowker space is through negating condition 3 or condition 4. Thus, given an infinite cardinal $\kappa$, a $\kappa$-Dowker space is a normal space $X$ that satisfies the following condition:

There exists a decreasing family $\left\{F_\alpha: \alpha<\kappa \right\}$ of closed subsets of $X$ with $\bigcap_{\alpha<\kappa} F_\alpha=\varnothing$ such that for every family $\left\{G_\alpha: \alpha<\kappa \right\}$ of open subsets of $X$ with $F_\alpha \subset G_\alpha$ for each $\alpha$, $\bigcap_{\alpha<\kappa} G_\alpha \ne \varnothing$.

The definition of $\kappa$-Dowker space is through negating condition 4. Of course, negating condition 3 would give an equivalent definition.

When $\kappa$ is the countably infinite cardinal $\omega$, a $\kappa$-Dowker space is simply the ordinary Dowker space constructed by M. E. Rudin [2]. Rudin generalized the construction of the ordinary Dowker space to obtain a $\kappa$-Dowker space for every infinite cardinal $\kappa$ [4]. The space that Rudin constructed in [4] would be a normal space $X$ such that condition 4 of Theorem 2 is violated. This means that the space $X$ would violate condition 7 in Theorem 2. Thus $X \times Y$ is not normal for every space $Y$ with $mt(Y)=\kappa$.

Here’s the solution of Morita’s first conjecture. Let $Y$ be a normal and non-discrete space. Determine the least cardinality $\kappa$ of a non-discrete subspace of $Y$. Obtain the $\kappa$-Dowker space $X$ as in [4]. Then $X \times Y$ is not normal according to the preceding paragraph.

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Remarks

Answering Morita’s first conjecture is a two-step approach. First, figure out what a generalized Dowker’s theorem should be. Then a $\kappa$-Dowker space is one that violates an appropriate condition in the generalized Dowker’s theorem. By violating the right condition in the theorem, we have a way to obtain non-normal product space needed in the answer. The second step is of course the proof of the existence of a space that violates the condition in the generalized Dowker’s theorem.

Figuring out the form of the generalized Dowker’s theorem took some work. It is more than just changing the countable infinite cardinal in Dowker’s theorem (Theorem 1 above) to an arbitrary infinite cardinal. This is because the conditions in Theorem 1 are unequal when the cardinality is changed to an uncountable one.

We take the cue from Rudin’s chapter on Dowker spaces [3]. In the last page of that chapter, Rudin pointed out the conditions that should go into a generalized Dowker’s theorem. However, the explanation of the relationship among the conditions is not clear. The previous post and this post are an attempt to sort out the conditions and fill in as much details as possible.

Rudin’s chapter did have the right condition for defining $\kappa$-Dowker space. It seems that prior to the writing of that chapter, there was some confusion on how to define a $\kappa$-Dowker space, i.e. a condition in the theorem the violation of which would give a $\kappa$-Dowker space. If the condition used is a stronger property, the violation may not yield enough information to get non-normal products. According to Diagram 1, condition 3 in Theorem 2 is the right one to use since it is the weakest condition and is down streamed from the conditions about normal product space. So the violation of condition 3 would answer Morita’s first conjecture.

We do not discuss the other step in the solution in any details. Any interested reader can review Rudin’s construction in [2] and [4]. The $\kappa$-Dowker space is an appropriate subspace of a product space with the box topology.

One interesting observation about the ordinary Dowker space (the one that violates a condition in Theorem 1) is that the product of any Dowker space and any space with a countable non-discrete subspace is not normal. This shows that Dowker space is badly non-productive with respect to normality. This fact is actually not obvious in the usual formulation of Dowker’s theorem (Theorem 1 above). What makes this more obvious in the direction $7 \Longrightarrow 3$ in Theorem 2. For the countably infinite case, $7 \Longrightarrow 3$ is essentially this: If $X \times Y$ is normal where $Y$ has a countable non-discrete subspace, then $X$ is not a Dowker space. Thus if the goal is to find a non-normal product space, a Dowker space should be one space to check.

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Loose Ends

In the course of working on the contents in this post and the previous post, there are some questions that we do not know how to answer and have not spent time to verify one way or the other. Possibly there are some loose ends to tie. They for the most parts are not open questions, but they should be interesting questions to consider.

For the $\kappa$-Dowker’s theorem (Theorem 2), one natural question is on the relative strengths of the conditions. It will be interesting to find out the implications not shown in Diagram 1. For example, for the three shrinking properties (conditions 2, 3 and 5), it is straightforward from definition that $2 \Longrightarrow 3$ and $5 \Longrightarrow 3$. The example of $X=\omega_1$ (the first uncountable ordinal) shows that $2 \not \Longrightarrow 5$ and hence $2 \not \Longrightarrow 1$. What about $3 \Longrightarrow 2$? In [5], Beslagic and Rudin showed that $3 \not \Longrightarrow 2$ using $\Diamond ^{++}$. A natural question would be: can there be ZFC example? Perhaps searching on more recent papers can yield some answers.

Another question is $5 \Longrightarrow 1$? The answer is no with the example being a Navy space – Example 7.6 in p. 194 [1]. The other two directions that have not been accounted for are: $7 \Longrightarrow 6$ and $3 \Longrightarrow 7$? We do not know the answer.

Another small question that we come across is about $X=\omega_1$ (the first uncountable ordinal). This is an example for showing $2 \not \Longrightarrow 5$. Thus condition 6 is false. Thus $X \times Y_{\omega_1}$ is not normal. Here $Y_{\omega_1}$ is simply the one-point Lindelofication of a discrete space of cardinality $\omega_1$. The question is: is condition 7 true for $X=\omega_1$? The product of $X=\omega_1$ and $Y_{\omega_1}$ (a space with minimal tightness $\omega_1$) is not normal. Is there a normal $X \times Y$ where $Y$ is another space with minimal tightness $\omega_1$?

Dowker’s theorem and $\kappa$-Dowker’s theorem show that finding a normal space that is not shrinking is not a simple matter. To find a normal space that is not countably shrinking took 20 years (1951 to 1971). For any uncountable $\kappa$, the $\kappa$-Dowker space that is based on the same construction of an ordinary Dowker space is also a space that is not $\kappa$-shrinking. With an uncountable $\kappa$, is the $\kappa$-Dowker space countably shrinking? This is not obvious one way or the other just from the definition of $\kappa$-Dowker space. Perhaps there is something obvious and we have not connected the dots. Perhaps we need to go into the definition of the $\kappa$-Dowker space in [4] to show that it is countably shrinking. The motivation is that we tried to find a normal space that is countably shrinking but not $\kappa$-shrinking for some uncountable $\kappa$. It seems that the $\kappa$-Dowker space in [4] is the natural candidate.

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Reference

1. Morita K., Nagata J.,Topics in General Topology, Elsevier Science Publishers, B. V., The Netherlands, 1989.
2. Rudin M. E., A Normal Space $X$ for which $X \times I$ is not Normal, Fund. Math., 73, 179-486, 1971. (link)
3. Rudin M. E., Dowker Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, (1984) 761-780.
4. Rudin M. E., $\kappa$-Dowker Spaces, Czechoslovak Mathematical Journal, 28, No.2, 324-326, 1978. (link)
5. Rudin M. E., Beslagic A.,Set-Theoretic Constructions of Non-Shrinking Open Covers, Topology Appl., 20, 167-177, 1985. (link)
6. Yasui Y., On the Characterization of the $\mathcal{B}$-Property by the Normality of Product Spaces, Topology and its Applications, 15, 323-326, 1983. (abstract and paper)
7. Yasui Y., Some Characterization of a $\mathcal{B}$-Property, TSUKUBA J. MATH., 10, No. 2, 243-247, 1986.

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$\copyright \ 2017 \text{ by Dan Ma}$

# Spaces with shrinking properties

Certain covering properties and separation properties allow open covers to shrink, e.g. paracompact spaces, normal spaces, and countably paracompact spaces. The shrinking property is also interesting on its own. This post gives a more in-depth discussion than the one in the previous post on countably paracompact spaces. After discussing shrinking spaces, we introduce three shrinking related properties. These properties show that there is a deep and delicate connection among shrinking properties and normality in products. This post is also a preparation for the next post on $\kappa$-Dowker space and Morita’s first conjecture.

All spaces under consideration are Hausdorff and normal or Hausdorff and regular (if not normal).

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Shrinking Spaces

Let $X$ be a space. Let $\mathcal{U}$ be an open cover of $X$. The open cover of $\mathcal{U}$ is said to be shrinkable if there is an open cover $\mathcal{V}=\left\{V(U): U \in \mathcal{U} \right\}$ of $X$ such that $\overline{V(U)} \subset U$ for each $U \in \mathcal{U}$. When this is the case, the open cover $\mathcal{V}$ is said to be a shrinking of $\mathcal{U}$. If an open cover is shrinkable, we also say that the open cover can be shrunk (or has a shrinking). Whenever an open cover has a shrinking, the shrinking is indexed by the open cover that is being shrunk. Thus if the original cover is indexed in a certain way, e.g. $\left\{U_\alpha: \alpha<\kappa \right\}$, then a shrinking has the same indexing, e.g. $\left\{V_\alpha: \alpha<\kappa \right\}$.

A space $X$ is a shrinking space if every open cover of $X$ is shrinkable. The property can also be broken up according to the cardinality of the open cover. Let $\kappa$ be a cardinal. A space $X$ is $\kappa$-shrinking if every open cover of cardinality $\le \kappa$ for $X$ is shrinkable. A space $X$ is countably shrinking if it is $\omega$-shrinking.

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Examples of Shrinking

Let’s look at a few situations where open covers can be shrunk either all the time or on a limited basis. For a normal space, certain covers can be shrunk as indicated by the following theorem.

Theorem 1
The following conditions are equivalent.

1. The space $X$ is normal.
2. Every point-finite open cover of $X$ is shrinkable.
3. Every locally finite open cover of $X$ is shrinkable.
4. Every finite open cover of $X$ is shrinkable.
5. Every two-element open cover of $X$ is shrinkable.

The hardest direction in the proof is $1 \Longrightarrow 2$, which is established in this previous post. The directions $2 \Longrightarrow 3 \Longrightarrow 4 \Longrightarrow 5$ are immediate. To see $5 \Longrightarrow 1$, let $H$ and $K$ be two disjoint closed subsets of $X$. By condition 5, the two-element open cover $\left\{X-H,X-K \right\}$ has a shrinking $\left\{U,V \right\}$. Then $\overline{U} \subset X-H$ and $\overline{V} \subset X-K$. As a result, $H \subset X-\overline{U}$ and $K \subset X-\overline{V}$. Since the open sets $U$ and $V$ cover the whole space, $X-\overline{U}$ and $X-\overline{V}$ are disjoint open sets. Thus $X$ is normal.

In a normal space, all finite open covers are shrinkable. In general, an infinite open cover of a normal space does not have to be shrinkable unless it is a point-finite or locally finite open cover.

The theorem of C. H. Dowker states that a normal space $X$ is countably paracompact if and only every countable open cover of $X$ is shrinkable if and only if the product space $X \times Y$ is normal for every compact metric space $Y$ if and only if the product space $X \times [0,1]$ is normal. The theorem is discussed here. A Dowker space is a normal space that violates the theorem. Thus any Dowker space has a countably infinite open cover that cannot be shrunk, or equivalently a normal space that forms a non-normal product with a compact metric space. Thus the notion of shrinking has a connection with normality in the product spaces. A Dowker space space was constructed by M. E. Rudin in ZFC [2]. So far Rudin’s example is essentially the only ZFC Dowker space. This goes to show that finding a normal space that is not countably shrinking is not a trivial matter.

Several facts can be derived easily from Theorem 1 and Dowker’s theorem. For clarity, they are called out as corollaries.

Corollary 2

• All shrinking spaces are normal.
• All shrinking spaces are normal and countably paracompact.
• Any normal and metacompact space is a shrinking space.

For the first corollary, if every open cover of a space can be shrunk, then all finite open covers can be shrunk and thus the space must be normal. As indicated above, Dowker’s theorem states that in a normal space, countably paracompactness is equivalent to countably shrinking. Thus any shrinking space is normal and countably paracompact.

Though an infinite open cover of a normal space may not be shrinkable, adding an appropriate covering property to any normal space will make it into a shrinking space. An easy way is through point-finite open covers. If every open cover has a point-finite open refinement (i.e. a metacompact space), then the point-finite open refinement can be shrunk (if the space is also normal). Thus the third corollary is established. Note that the metacompact is not the best possible result. For example, it is known that any normal and submetacompact space is a shrinking space – see Theorem 6.2 of [1].

In paracompact spaces, all open covers can be shrunk. One way to see this is through Corollary 2. Any paracompact space is normal and metacompact. It is also informative to look at the following characterization of paracompact spaces.

Theorem 3
A space $X$ is paracompact if and only if every open cover $\left\{U_\alpha: \alpha<\kappa \right\}$ of $X$ has a locally finite open refinement $\left\{V_\alpha: \alpha<\kappa \right\}$ such that $\overline{V_\alpha} \subset U_\alpha$ for each $\alpha$.

A proof can be found here. Thus every open cover of a paracompact space can be shrunk by a locally finite shrinking. To summarize, we have discussed the following implications.

Diagram 1

\displaystyle \begin{aligned} \text{Paracompact} \Longrightarrow & \text{ Normal + Metacompact} \\&\ \ \ \ \ \ \Big \Downarrow \\&\text{ Shrinking} \\&\ \ \ \ \ \ \Big \Downarrow \\& \text{ Normal + Countably Paracompact} \\&\ \ \ \ \ \ \Big \Downarrow \\& \text{ Normal} \end{aligned}

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Three Shrinking Related Properties

None of the implications in Diagram 1 can be reversed. The last implication in the diagram cannot be reversed due to Rudin’es Dowker space. One natural example to look for would be spaces that are normal and countably paracompact but fail in shrinking at some uncountable cardinal. As indicated by the the theorem of C. H, Dowker, the notion of shrinking is intimately connected to normality in product spaces $X \times Y$. To further investigate, consider the following three properties.

Let $X$ be a space. Let $\kappa$ be an infinite cardinal. Consider the following three properties.

The space $X$ is $\kappa$-shrinking if and only if any open cover of cardinality $\le \kappa$ for the space $X$ is shrinkable, i.e. the following condition holds.

For each open cover $\left\{U_\alpha: \alpha<\kappa \right\}$ of $X$, there exists an open cover $\left\{V_\alpha: \alpha<\kappa \right\}$ such that $\overline{V_\alpha} \subset U_\alpha$ for each $\alpha<\kappa$.

The space $X$ has Property $\mathcal{D}(\kappa)$ if and only if every increasing open cover of cardinality $\le \kappa$ for the space $X$ is shrinkable, i.e. the following holds.

For each increasing open cover $\left\{U_\alpha: \alpha<\kappa \right\}$ of $X$, there exists an open cover $\left\{V_\alpha: \alpha<\kappa \right\}$ such that $\overline{V_\alpha} \subset U_\alpha$ for each $\alpha<\kappa$.

The space $X$ has Property $\mathcal{B}(\kappa)$ if and only if the following holds.

For each increasing open cover $\left\{U_\alpha: \alpha<\kappa \right\}$ of $X$, there exists an increasing open cover $\left\{V_\alpha: \alpha<\kappa \right\}$ such that $\overline{V_\alpha} \subset U_\alpha$ for each $\alpha<\kappa$.

A family $\left\{A_\alpha: \alpha<\kappa \right\}$ is increasing if $A_\alpha \subset A_\beta$ for any $\alpha<\beta<\kappa$. It is decreasing if $A_\beta \subset A_\alpha$ for any $\alpha<\beta<\kappa$.

In general, any space that is $\kappa$-shrinking for all cardinals $\kappa$ is a shrinking space as defined earlier. Any space that has property $\mathcal{D}(\kappa)$ for all cardinals $\kappa$ is said to have property $\mathcal{D}$. Any space that has property $\mathcal{B}(\kappa)$ for all cardinals $\kappa$ is said to have property $\mathcal{B}$.

The first property $\kappa$-shrinking is simply the shrinking property for open covers of cardinality $\le \kappa$. The property $\mathcal{D}(\kappa)$ is $\kappa$-shrinking with the additional requirement that the open covers to be shrunk must be increasing. It is clear that $\kappa$-shrinking implies property $\mathcal{D}(\kappa)$. The property $\mathcal{B}(\kappa)$ appears to be similar to $\mathcal{D}(\kappa)$ except that $\mathcal{B}(\kappa)$ has the additional requirement that the shrinking is also increasing. As a result $\mathcal{B}(\kappa)$ implies $\mathcal{D}(\kappa)$. The following diagram shows the implications.

Diagram 2

$\displaystyle \begin{array}{ccccc} \kappa \text{-Shrinking} &\text{ } & \not \longrightarrow & \text{ } & \text{Property } \mathcal{B}(\kappa) \\ \text{ } & \searrow & \text{ } & \swarrow & \text{ } \\ \text{ } &\text{ } & \text{Property } \mathcal{D}(\kappa) & \text{ } & \text{ } \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ \end{array}$

The implications in Diagram 2 are immediate. An example is given below showing that $\omega_1$-shrinking does not imply property $\mathcal{B}(\omega_1)$. If $\kappa=\omega$, then all three properties are equivalent in normal spaces, as displayed in the following diagram. The proof is in Theorem 5.

Diagram 3

$\displaystyle \begin{array}{ccccc} \omega \text{-Shrinking} &\text{ } & \longrightarrow & \text{ } & \text{Property } \mathcal{B}(\omega) \\ \text{ } & \nwarrow & \text{ } & \swarrow & \text{ } \\ \text{ } &\text{ } & \text{Property } \mathcal{D}(\omega) & \text{ } & \text{ } \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ \end{array}$

The property $\mathcal{D}(\kappa)$ has a dual statement in terms of decreasing closed sets. The following theorem gives the dual statement.

Theorem 4
Let $X$ be a normal space. Let $\kappa$ be an infinite cardinal. The following two properties are equivalent.

• The space $X$ has property $\mathcal{D}(\kappa)$.
• For each decreasing family $\left\{F_\alpha: \alpha<\kappa \right\}$ of closed subsets of $X$ such that $\bigcap_{\alpha<\kappa} F_\alpha=\varnothing$, there exists a family $\left\{G_\alpha: \alpha<\kappa \right\}$ of open subsets of $X$ such that $\bigcap_{\alpha<\kappa} G_\alpha=\varnothing$ and $F_\alpha \subset G_\alpha$ for each $\alpha<\kappa$.

First bullet implies second bullet
Let $\left\{F_\alpha: \alpha<\kappa \right\}$ be a decreasing family of closed subsets of $X$ with empty intersection. Then $\left\{U_\alpha: \alpha<\kappa \right\}$ is an increasing family of open subsets of $X$ where $U_\alpha=X-F_\alpha$. Let $\left\{V_\alpha: \alpha<\kappa \right\}$ be an open cover of $X$ such that $\overline{V_\alpha} \subset U_\alpha$ for each $\alpha$. Then $\left\{G_\alpha: \alpha<\kappa \right\}$ where $G_\alpha=X-\overline{V_\alpha}$ is the needed open expansion.

Second bullet implies first bullet
Let $\left\{U_\alpha: \alpha<\kappa \right\}$ be an increasing open cover of $X$. Then $\left\{F_\alpha: \alpha<\kappa \right\}$ is a decreasing family of closed subsets of $X$ where $F_\alpha=X-U_\alpha$. Note that $\bigcap_{\alpha<\kappa} F_\alpha=\varnothing$. Let $\left\{G_\alpha: \alpha<\kappa \right\}$ be a family of open subsets of $X$ such that $\bigcap_{\alpha<\kappa} G_\alpha=\varnothing$ and $F_\alpha \subset G_\alpha$ for each $\alpha$. For each $\alpha$, there is open set $W_\alpha$ such that $F_\alpha \subset W_\alpha \subset \overline{W_\alpha} \subset G_\alpha$ since $X$ is normal. For each $\alpha$, let $V_\alpha=X-\overline{W_\alpha}$. Then $\left\{V_\alpha: \alpha<\kappa \right\}$ is a family of open subsets of $X$ required by the first bullet. It is a cover because $\bigcap_{\alpha<\kappa} \overline{W_\alpha}=\varnothing$. To show $\overline{V_\alpha} \subset U_\alpha$, let $x \in \overline{V_\alpha}$ such that $x \notin U_\alpha$. Then $x \in W_\alpha$. Since $x \in \overline{V_\alpha}$ and $W_\alpha$ is open, $W_\alpha \cap V_\alpha \ne \varnothing$. Let $y \in W_\alpha \cap V_\alpha$. Since $y \in V_\alpha$, $y \notin \overline{W_\alpha}$, which means $y \notin W_\alpha$, a contradiction. Thus $\overline{V_\alpha} \subset U_\alpha$.

Now we show that the three properties in Diagram 3 are equivalent.

Theorem 5
Let $X$ be a normal space. Then the following implications hold.
$\omega$-shrinking $\Longrightarrow$ Property $\mathcal{B}(\omega)$ $\Longrightarrow$ Property $\mathcal{D}(\omega)$ $\Longrightarrow$ $\omega$-shrinking

Proof of Theorem 5
$\omega$-shrinking $\Longrightarrow$ Property $\mathcal{B}(\omega)$
Suppose that $X$ is $\omega$-shrinking. By Dowker’s theorem, $X \times (\omega+1)$ is a normal space. We can think of $\omega+1$ as a convergent sequence with $\omega$ as the limit point. Let $\left\{U_n:n=0,1,2,\cdots \right\}$ be an increasing open cover of $X$. Define $H$ and $K$ as follows:

$H=\cup \left\{(X-U_n) \times \left\{n \right\}: n=0,1,2,\cdots \right\}$

$K=X \times \left\{\omega \right\}$

It is straightforward to verify that $H$ and $K$ are disjoint closed subsets of $X \times (\omega+1)$. By normality, let $V$ and $W$ be disjoint open subsets of $X \times (\omega+1)$ such that $H \subset W$ and $K \subset V$. For each integer $n=0,1,2,\cdots$, define $V_n$ as follows:

$V_n=\left\{x \in X: \exists \ \text{open } O \subset X \text{ such that } x \in O \text{ and } O \times [n, \omega] \subset V \right\}$

The set $[n, \omega]$ consists of all integers $\ge n$ and the limit point $\omega$. From the way the sets $V_n$ are defined, $\left\{V_n:n=0,1,2,\cdots \right\}$ is an increasing open cover of $X$. The remaining thing to show is that $\overline{V_n} \subset U_n$ for each $n$. Suppose that $x \in \overline{V_n}$ and $x \notin U_n$. Then $(x,n) \in H$ by definition of $H$. There exists an open set $E \times \left\{n \right\}$ such that $(x,n) \in E \times \left\{n \right\}$ and $(E \times \left\{n \right\}) \cap V=\varnothing$. Since $E$ is an open set containing $x$, $E \cap V_n \ne \varnothing$. Let $y \in E \cap V_n$. By definition of $V_n$, there is some open set $O$ such that $y \in O$ and $O \times [n, \omega] \subset V$, a contradiction since $(E \cap O) \times \left\{n \right\}$ is supposed to miss $V$. Thus $\overline{V_n} \subset U_n$ for all integers $n$.

The direction Property $\mathcal{B}(\omega)$ $\Longrightarrow$ Property $\mathcal{D}(\omega)$ is immediate.

Property $\mathcal{D}(\omega)$ $\Longrightarrow$ $\omega$-shrinking
Consider the dual condition of $\mathcal{D}(\omega)$ in Theorem 4, which is equivalent to $\omega$-shrinking according to Dowker’s theorem. $\square$

Remarks
The direction $\omega$-shrinking $\Longrightarrow$ Property $\mathcal{B}(\omega)$ is true because $\omega$-shrinking is equivalent to the normality in the product $X \times (\omega+1)$. The same is not true when $\kappa$ becomes an uncountable cardinal. We now show that $\kappa$-shrinking does not imply $\mathcal{B}(\kappa)$ in general.

Example 1
The space $X=\omega_1$ is the set of all ordinals less than $\omega_1$ with the ordered topology. Since it is a linearly ordered space, it is a shrinking space. Thus in particular it is $\omega_1$-shrinking. To show that $X$ does not have property $\mathcal{B}(\omega_1)$, consider the increasing open cover $\left\{U_\alpha: \alpha<\omega_1 \right\}$ where $U_\alpha=[0,\alpha)$ for each $\alpha<\omega_1$. Here $[0,\alpha)$ consists of all ordinals less than $\alpha$. Suppose $X$ has property $\mathcal{B}(\omega_1)$. Then let $\left\{V_\alpha: \alpha<\omega_1 \right\}$ be an increasing open cover of $X$ such that $\overline{V_\alpha} \subset U_\alpha$ for each $\alpha$.

Let $L$ be the set of all limit ordinals in $X$. For each $\alpha \in L$, $\alpha \notin U_\alpha$ and thus $\alpha \notin \overline{V_\alpha}$. Thus there exists a countable ordinal $f(\alpha)<\alpha$ such that $(f(\alpha),\alpha]$ misses points in $\overline{V_\alpha}$. Thus the map $f: L \rightarrow \omega_1$ is a pressing down map. By the pressing down lemma, there exists some $\alpha<\omega_1$ such that $S=f^{-1}(\alpha)$ is a stationary set in $\omega_1$, which means that $S$ intersects with every closed and unbounded subset of $X=\omega_1$. This means that for each $\gamma>\alpha$, $(\alpha, \gamma]$ would miss $\overline{V_\gamma}$. This means that for each $\gamma>\alpha$, $\overline{V_\gamma} \subset [0,\alpha]$. As a result $\left\{V_\alpha: \alpha<\omega_1 \right\}$ would not be a cover of $X$, a contradiction. So $X$ does not have property $\mathcal{B}(\omega_1)$. $\square$

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Property $\mathcal{B}(\kappa)$

Of the three properties discussed in the above section, we would like to single out property $\mathcal{B}(\kappa)$. This property has a connection with normality in the product $X \times Y$ (see Theorem 7). First, we prove a lemma that is used in proving Theorem 7.

Lemma 6
Show that the property $\mathcal{B}(\kappa)$ is hereditary with respect to closed subsets.

Proof of Lemma 6
Let $X$ be a space with property $\mathcal{B}(\kappa)$. Let $A$ be a closed subspace of $X$. Let $\left\{U_\alpha \subset A: \alpha<\kappa \right\}$ be an increasing open cover of $A$. For each $\alpha$, let $W_\alpha$ be an open subset of $X$ such that $U_\alpha=W_\alpha \cap A$. Since the open sets $U_\alpha$ are increasing, the open sets $W_\alpha$ can be chosen inductively such that $W_\alpha \supset W_\gamma$ for all $\gamma<\alpha$. This will ensure that $W_\alpha$ will form an increasing cover.

Then $\left\{W_\alpha^* \subset X: \alpha<\kappa \right\}$ is an increasing open cover of $X$ where $W_\alpha^*=W_\alpha \cup (X-A)$. By property $\mathcal{B}(\kappa)$, let $\left\{E_\alpha \subset X: \alpha<\kappa \right\}$ be an increasing open cover of $X$ such that $\overline{E_\alpha} \subset W_\alpha^*$. For each $\alpha$, let $V_\alpha=E_\alpha \cap A$. It can be readily verified that $\left\{V_\alpha \subset A: \alpha<\kappa \right\}$ is an increasing open cover of $A$. Furthermore, $\overline{V_\alpha} \subset U_\alpha$ for each $\alpha$ (closure taken in $A$). $\square$

Let $\kappa$ be an infinite cardinal. Let $D_\kappa=\left\{d_\alpha: \alpha<\kappa \right\}$ be a discrete space of cardinality $\kappa$. Let $p$ be a point not in $D_\kappa$. Let $Y_\kappa=D_\kappa \cup \left\{p \right\}$. Define a topology on $Y_\kappa$ by letting $D_\kappa$ be discrete and by letting open neighborhood of $p$ be of the form $\left\{p \right\} \cup E$ where $E \subset D_\kappa$ and $D_\kappa-E$ has cardinality less than $\kappa$. Note the similarity between $Y_\kappa$ and the convergent sequence $\omega+1$ in the proof of Theorem 5.

Theorem 7
Let $X$ be a normal space. Then the product space $X \times Y_\kappa$ is normal if and only if $X$ has property $\mathcal{B}(\kappa)$.

Remarks
The property $\mathcal{B}(\kappa)$ involves the shrinking of any increasing open cover with the added property that the shrinking is also increasing. The increasing shrinking is just what is needed to show that disjoint closed subsets of the product space can be separated.

Notations
Let’s set some notations that are useful in proving Theorem 7.

• The set $[d_\alpha,p]$ is an open set in $Y_\kappa$ containing the point $p$ and is defined as follows.
• $[d_\alpha,p]=\left\{d_\beta: \alpha \le \beta<\kappa \right\} \cup \left\{p \right\}$.
• For any two disjoint closed subsets $H$ and $K$ of the product space $X \times Y_\kappa$, define the following sets.
• For each $\alpha<\kappa$, let $H_\alpha=H \cap (X \times \left\{d_\alpha \right\})$ and $K_\alpha=K \cap (X \times \left\{d_\alpha \right\})$.
• Let $H_p=H \cap (X \times \left\{p \right\})$ and $K_p=K \cap (X \times \left\{p \right\})$.
• For each $\alpha<\kappa$, choose open $O_\alpha \subset X$ such that $G_\alpha=O_\alpha \times \left\{d_\alpha \right\}$, $H_\alpha \subset G_\alpha$ and $\overline{G_\alpha} \cap K_\alpha=\varnothing$ (due to normality of $X$).
• Choose open $O_p \subset X$ such that $G_p=O_p \times \left\{p \right\}$, $H_p \subset G_p$ and $\overline{G_p} \cap K_p=\varnothing$ (due to normality of $X$).

Proof of Theorem 7
Suppose that $X$ has property $\mathcal{B}(\kappa)$. Let $H$ and $K$ be two disjoint closed sets of $X \times Y_\kappa$. Consider the following cases based on the locations of the closed sets $H$ and $K$.

Case 1. $H \subset X \times D_\kappa$ and $K \subset X \times D_\kappa$.
Case 2a. $H=X \times \left\{p\right\}$
Case 2b. Exactly one of $H$ and $K$ intersect the set $X \times \left\{p\right\}$.
Case 3. Both $H$ and $K$ intersect the set $X \times \left\{p\right\}$.

Remarks
Case 1 is easy. Case 2a is the pivotal case. Case 2b and Case 3 use a similar idea. The result in Theorem 7 is found in [1] (Theorem 6.9 in p. 189) and [4]. The authors in these two sources claimed that Case 2a is the only case that matters, citing a lemma in another source. The lemma was not stated in these two sources and the source for the lemma is a PhD dissertation that is not readily available. Case 3 essentially uses the same idea but it has enough differences. For the sake of completeness, we work out all the cases. Case 3 applies property $\mathcal{B}(\kappa)$ twice. Despite the complicated notations, the essential idea is quite simple. If any reader finds the proof too long, just understand Case 2a and then get the gist of how the idea is applied in Case 2b and Case 3.

Case 1.
$H \subset X \times D_\kappa$ and $K \subset X \times D_\kappa$.

Let $M =\bigcup_{\alpha<\kappa} G_\alpha$. It is clear that $H \subset M$ and $\overline{M} \cap K=\varnothing$.

Case 2a.
Assume that $H=X \times \left\{p\right\}$. We now proceed to separate $H$ and $K$ with disjoint open sets. For each $\alpha<\kappa$, define $U_\alpha$ as follows:

$U_\alpha=\cup \left\{O \subset X: O \text{ is open such that } (O \times [d_\alpha,p]) \cap K =\varnothing \right\}$

Then $\left\{U_\alpha: \alpha<\kappa \right\}$ is an increasing open cover of $X$. By property $\mathcal{B}(\kappa)$, there is an increasing open cover $\mathcal{V}=\left\{V_\alpha: \alpha<\kappa \right\}$ of $X$ such that $\overline{V_\alpha} \subset U_\alpha$ for each $\alpha$. The shrinking $\mathcal{V}$ allows us to define an open set $G$ such that $H \subset G$ and $\overline{G} \cap K=\varnothing$.

Let $G=\cup \left\{V_\alpha \times [d_\alpha,p]: \alpha<\kappa \right\}$. It is clear that $H \subset G$. Next, we show that $\overline{G} \cap K=\varnothing$. Suppose that $(x,d_\alpha) \in K$. Then $(x,d_\alpha) \notin U_\alpha \times [d_\alpha,p]$. As a result, $(x,d_\alpha) \notin \overline{V_\alpha} \times [d_\alpha,p]$. Let $O \subset X$ be open such that $x \in O$ and $(O \times \left\{d_\alpha \right\}) \cap (\overline{V_\alpha} \times [d_\alpha,p])=\varnothing$. Since $V_\beta \subset V_\alpha$ for all $\beta<\alpha$, it follows that $(O \times \left\{d_\alpha \right\}) \cap (V_\beta \times [d_\beta,p])=\varnothing$ for all $\beta < \alpha$. It is clear that $(O \times \left\{d_\alpha \right\}) \cap (V_\gamma \times [d_\gamma,p])=\varnothing$ for all $\gamma>\alpha$. What has been shown is that there is an open set containing the point $(x,d_\alpha)$ that contains no point of $G$. This means that $(x,d_\alpha) \notin \overline{G}$. We have established that $\overline{G} \cap K=\varnothing$.

Case 2b.
Exactly one of $H$ and $K$ intersect the set $X \times \left\{p\right\}$. We assume that $H$ is the set that intersects the set $X \times \left\{p\right\}$. The only difference between Case 2b and Case 2a is that there can be points of $H$ outside of $X \times \left\{p\right\}$ in Case 2b.

Now proceed as in Case 2a. Obtain the open cover $\left\{U_\alpha: \alpha<\kappa \right\}$, the open cover $\left\{V_\alpha: \alpha<\kappa \right\}$ and the open set $G$ as in Case 2a. Let $M=G \cup (\bigcup_{\alpha<\kappa} G_\alpha)$. It is clear that $H \subset M$. We claim that $\overline{M} \cap K=\varnothing$. Suppose that $(x,d_\gamma) \in K$. Since $\overline{G} \cap K=\varnothing$ (as in Case 2a), there exists open set $W=O \times \left\{ d_\gamma \right\}$ such that $(x,d_\gamma) \in W$ and $W \cap \overline{G}=\varnothing$. There also exists open $W_1 \subset W$ such that $(x,d_\gamma) \in W_1$ and $W_1 \cap \overline{G_\gamma}=\varnothing$. It is clear that $W_1 \cap G_\beta=\varnothing$ for all $\beta \ne \gamma$. This means that $W_1$ is an open set containing the point $(x,d_\gamma)$ such that $W_1$ misses the open set $M$. Thus $\overline{M} \cap K=\varnothing$.

Case 3.
Both $H$ and $K$ intersect the set $X \times \left\{p\right\}$.

Now project $H_p$ and $K_p$ onto the space $X$.

$H_p^*=\left\{x \in X: (x,p) \in H_p \right\}$

$K_p^*=\left\{x \in X: (x,p) \in K_p \right\}$

Note that $H_p^*$ is simply the copy of $H_p$ and $K_p^*$ is the copy of $K_p$ in $X$. Since $X$ is normal, choose disjoint open sets $E_1$ and $E_1$ such that $H_p^* \subset E_1$ and $K_p^* \subset E_2$.

Let $A_1=\overline{E_1}$ and $B_1=X-K_p^*$. Let $A_2=\overline{E_2}$ and $B_2=X-H_p^*$. Note that $A_1$ is closed in $X$, $B_1$ is open in $X$ and $A_1 \subset B_1$. Similarly $A_2$ is closed in $X$, $B_2$ is open in $X$ and $A_2 \subset B_2$.

We now define two increasing open covers using property $\mathcal{B}(\kappa)$. Define $U_{\alpha,1}$ and $T_{\alpha,1}$ and $U_{\alpha,2}$ and $T_{\alpha,2}$ as follows:

$U_{\alpha,1}=\cup \left\{O \subset B_1: O \text{ is open such that } (O \times [d_\alpha,p]) \cap K =\varnothing \right\}$

$T_{\alpha,1}=U_{\alpha,1} \cap A_1$

$U_{\alpha,2}=\cup \left\{O \subset B_2: O \text{ is open such that } (O \times [d_\alpha,p]) \cap H =\varnothing \right\}$

$T_{\alpha,2}=U_{\alpha,2} \cap A_2$

The open cover $\mathcal{T}_1=\left\{T_{\alpha,1}: \alpha<\kappa \right\}$ is an increasing open cover of $A_1$. The open cover $\mathcal{T}_2=\left\{T_{\alpha,2}: \alpha<\kappa \right\}$ is an increasing open cover of $A_2$.By property $\mathcal{B}(\kappa)$ of $A_1$ and $A_2$, both covers have the following as shrinking (by Lemma 6). The two shrinkings are:

$\mathcal{V}_1=\left\{V_{\alpha,1} \subset A_1: \alpha<\kappa \right\}$

$\mathcal{V}_2=\left\{V_{\alpha,2} \subset A_2: \alpha<\kappa \right\}$

such that

$\overline{V_{\alpha,1}} \subset T_{\alpha,1}$

$\overline{V_{\alpha,2}} \subset T_{\alpha,2}$

for each $\alpha<\kappa$ and such that both $\mathcal{V}_1$ and $\mathcal{V}_2$ are increasing open covers. Note that the closure $\overline{V_{\alpha,1}}$ is taken in $A_1$ and the closure $\overline{V_{\alpha,2}}$ is taken in $A_2$.

For each $\alpha$, let $W_{\alpha,1}$ be the interior of $V_{\alpha,1}$ and $W_{\alpha,2}$ be the interior of $V_{\alpha,2}$ (with respect to $X$). Note that $W_{\alpha,1}$ is meaningful since $V_{\alpha,1}$ is a subset of the closure of the open set $E_1$. Similar observation for $W_{\alpha,2}$. To make the rest of the argument easier to see, note the following fact about $W_{\alpha,1}$ and $W_{\alpha,2}$.

$\overline{W_{\alpha,1}} \subset \overline{V_{\alpha,1}} \subset T_{\alpha,1} \subset U_{\alpha,1}$ (closure with respect to $X$)

$\overline{W_{\alpha,2}} \subset \overline{V_{\alpha,2}} \subset T_{\alpha,2} \subset U_{\alpha,2}$ (closure with respect to $X$)

For each $\alpha<\kappa$, choose open set $O_\alpha \subset X$ such that

$L_\alpha=O_\alpha \times \left\{d_\alpha \right\}$

$H_\alpha \subset L_\alpha$

$\overline{L_\alpha} \cap K_\alpha=\varnothing$

$L_\alpha \cap (\overline{W_{\alpha,2}} \times [d_\alpha,p])=\varnothing$

The last point is possible because $U_{\alpha,2} \times [d_\alpha,p]$ misses $H$ and $\overline{W_{\alpha,2}} \subset U_{\alpha,2}$. Define the open sets $G$ and $M$ as follows:

$G=\cup \left\{W_{\alpha,1} \times [d_\alpha,p]: \alpha<\kappa \right\}$

$M=G \cup (\bigcup_{\alpha<\kappa} L_\alpha)$

It is clear that $H \subset M$. We claim that $\overline{M} \cap K=\varnothing$. To this end, we show that if $(x,y) \in K$, then $(x,y) \notin \overline{M}$. If $(x,y) \in K$, then either $(x,y)=(x,d_\gamma)$ for some $\gamma$ or $(x,y)=(x,p)$.

Let $(x,d_\gamma) \in K$. Note that $(x,d_\gamma) \notin U_{\gamma,1} \times [d_\gamma,p]$. Since $\overline{W_{\gamma,1}} \subset \overline{V_{\gamma,1}} \subset T_{\gamma,1} \subset U_{\gamma,1}$, $(x,d_\gamma) \notin \overline{W_{\gamma,1}} \times [d_\gamma,p]$. Choose an open set $O \subset X$ such that $x \in O$ and $C=O \times \left\{d_\gamma \right\}$ misses $\overline{W_{\gamma,1}} \times [d_\gamma,p]$. Note that $C$ misses $W_{\beta,1} \times [d_\beta,p]$ for all $\beta<\gamma$ since $W_{\beta,1} \subset W_{\gamma,1}$ for all $\beta<\gamma$. It is clear that $C$ misses $W_{\beta,1} \times [d_\beta,p]$ for all $\beta>\gamma$.

We can also choose open $C_1 \subset C$ such that $(x,d_\gamma) \in C_1$ and $C_1$ misses $\overline{L_\gamma}$. It is clear that $C_1$ misses $L_\beta$ for all $\beta \ne \gamma$. Thus there is an open set $C_1$ containing the point $(x,d_\gamma)$ such that $C_1$ contains no point of $M$.

Let $(x,p) \in K$. First we find an open set $Q$ containing $(x,p)$ such that $Q$ misses $G$. From the way the open sets $U_{\alpha,1}$ are defined, it follows that $(x,p) \notin \overline{W_{\alpha,1}} \times [d_\alpha,p]$ for all $\alpha$. Furthermore $W_{\alpha,1} \subset \overline{A_1}$. Thus $Q=(X-\overline{A_1}) \times Y_\kappa$ is the desired open set. On the other hand, there exists $\alpha<\kappa$ such that $x \in W_{\alpha,2}$. Note that $L_\gamma$ are chosen so that $(W_{\gamma,2} \times [d_\gamma,p]) \cap L_\gamma=\varnothing$ for all $\gamma$. Since $W_{\alpha,2} \subset W_{\beta,2}$ for all $\beta \ge \alpha$, $(W_{\alpha,2} \times [d_\alpha,p]) \cap L_\beta=\varnothing$ for all $\beta \ge \alpha$. Thus the open set $W_{\alpha,2} \times [d_\alpha,p]$ contains no points of $L_\gamma$ for any $\gamma$. Then the open set $Q \cap (W_{\alpha,2} \times [d_\alpha,p])$ contains no point of $M$. This means that $(x,p) \notin \overline{M}$. Thus $\overline{M} \cap K=\varnothing$.

In each of the four cases (1, 2a, 2b and 3), there exists an open set $M \subset X \times Y_\kappa$ such that $H \subset M$ and $\overline{M} \cap K=\varnothing$. This completes the proof that $X \times Y_\kappa$ is normal assuming that $X$ has property $\mathcal{B}(\kappa)$.

Now the other direction. Suppose that $X \times Y_\kappa$ is normal. Then it can be shown that $X$ has property $\mathcal{B}(\kappa)$. The proof is similar to the proof for $\omega$-shrinking $\Longrightarrow$ Property $\mathcal{B}(\omega)$ in Theorem 5. $\square$

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Reference

1. Morita K., Nagata J.,Topics in General Topology, Elsevier Science Publishers, B. V., The Netherlands, 1989.
2. Rudin M. E., A Normal Space $X$ for which $X \times I$ is not Normal, Fund. Math., 73, 179-486, 1971. (link)
3. Rudin M. E., Dowker Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, (1984) 761-780.
4. Yasui Y., On the Characterization of the $\mathcal{B}$-Property by the Normality of Product Spaces, Topology and its Applications, 15, 323-326, 1983. (abstract and paper)
5. Yasui Y., Some Characterization of a $\mathcal{B}$-Property, TSUKUBA J. MATH., 10, No. 2, 243-247, 1986.

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$\copyright \ 2017 \text{ by Dan Ma}$